Given the head of a singly linked list, reverse the list, and return the reversed list.
- Initialize three pointers prev as NULL, curr as head and next as NULL.
- Iterate through the linked list. In loop, do following.
- Before changing next of current, store next node next = curr->next
- Now change next of current This is where actual reversing happens curr->next = prev
- Move prev and curr one step forward prev = curr curr = next
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode *prev = NULL,*next=NULL,*curr=head;
while(curr!=NULL){
next=curr->next;
curr->next=prev;
prev=curr;
curr=next;
}
return prev;
}
};
- Divide the list in two parts - first node and rest of the linked list.
- Call reverse for the rest of the linked list.
- Link rest to first.
- Fix head pointer.
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==NULL || head->next==NULL) return head;
ListNode *rest = reverseList(head->next);
head->next->next=head;
head->next=NULL;
return rest;
}
};